Lamina

**JaysFan31** · July 30th 2006, 06:30 PM

Any help on this problem would be greatly appreciated. It deals with multiple integration:

Find the centre of mass (x,y) of the lamina which occupies the region:
36 less than or equal to (x^2) + (y^2) less than or equal to 49
y greater or equal to 0
if the density at any point is proportional to the distance from the origin.

Thanks in advance.

Mike

**ThePerfectHacker** · July 30th 2006, 06:50 PM

Originally Posted by JaysFan31

Any help on this problem would be greatly appreciated. It deals with multiple integration:

Find the centre of mass (x,y) of the lamina which occupies the region:
36 less than or equal to (x^2) + (y^2) less than or equal to 49
y greater or equal to 0
if the density at any point is proportional to the distance from the origin.

Thanks in advance.

Mike

Below is the diamgram of the region.
In order to solve this problem you need to find the countinous density function. Which you say is proportional to the distance. I take this to mean that you are saying directly proportional right? Then you have,
$\frac{\mbox{density}}{\mbox{distance}}=\kappa$ for some constant of proportionality.
In mathematical terms,
$\frac{\delta(x,y)}{\sqrt{x^2+y^2}}=\kappa$
Thus,
$\delta(x,y)=\kappa \sqrt{x^2+y^2}$

In order to find the center of mass you first need to find the mass. Which in this case is,
$\int_R \int \delta(x,y) dA$
Thus,
$\int_R \int \kappa \sqrt{x^2+y^2}dA$
I believe it is easier by converting to polar cordinates.
Thus,
$\int_0^{2\pi} \int_6^7 \kappa r^2 dr \, d\theta$
Thus,
$\int_0^{2\pi} \frac{1}{3}\kappa r^3 |_6^7= 84.\bar 6\pi \kappa$ --->Mass
----------
I will continue later.

**CaptainBlack** · July 30th 2006, 08:52 PM

Originally Posted by JaysFan31

Any help on this problem would be greatly appreciated. It deals with multiple integration:

Find the centre of mass (x,y) of the lamina which occupies the region:
36 less than or equal to (x^2) + (y^2) less than or equal to 49
y greater or equal to 0
if the density at any point is proportional to the distance from the origin.

Thanks in advance.

Mike

I will rewrite the definition of the region more clearly for you:

$ 36 \le x^2+y^2 \le 49, \ \ y \ge 0 $ ,

which is the top half of the shaded region in hacker's diagram.

RonL

(If the region had corresponded to hackers shaded region the centre
of mass would of course have been at $(0,0)$ by symmetry)

**CaptainBlack** · July 31st 2006, 04:35 AM

Originally Posted by JaysFan31

Any help on this problem would be greatly appreciated. It deals with multiple integration:

Find the centre of mass (x,y) of the lamina which occupies the region:
36 less than or equal to (x^2) + (y^2) less than or equal to 49
y greater or equal to 0
if the density at any point is proportional to the distance from the origin.

Thanks in advance.

Mike

Let the density $\rho(r)=\kappa r$ , theb=n the mass of the
lamina is:

$ M=\int_{\theta=0}^{\pi} \int_{r=6}^7 \rho(r)\ r\ dr d\theta$ $= \int_{\theta=0}^{\pi} \int_{r=6}^7 \kappa\ r^2dr\ d\theta $

Now the centre of mass is:

$ \bold{R}=\frac{1}{M} \int_A \rho(r)\ \bold{r} dA $

where $\bold{r}=r(\cos(\theta),\sin(\theta))$ .

So:

$ \bold{R}=\frac{1}{M}\int_{\theta=0}^{\pi} \int_{r=6}^7 \kappa\ r^3\ (\cos(\theta),\sin(\theta)) dr\ d\theta $ .

You should be able to complete these integrals yourself.

RonL

**ThePerfectHacker** · July 31st 2006, 01:26 PM

Originally Posted by CaptainBlank

Now the centre of mass is:

$ \bold{R}=\frac{1}{M} \int_A \rho(r)\ \bold{r} dA $

Is that a line integral? You can calculate center of mass with line integral?

**CaptainBlack** · July 31st 2006, 09:25 PM

Originally Posted by ThePerfectHacker

Is that a line integral? You can calculate center of mass with line integral?

It's an area integral.

**JaysFan31** · August 1st 2006, 08:56 AM

OK. I was able to get the centre of mass for x (it equals zero). How does the centre of mass for y differ? It does not equal zero. What changes in the integration process?

Thanks.

Mike

**CaptainBlack** · August 1st 2006, 09:55 AM

Originally Posted by JaysFan31

OK. I was able to get the centre of mass for x (it equals zero). How does the centre of mass for y differ? It does not equal zero. What changes in the integration process?

Thanks.

Mike

What changes is that $\sin(\theta)$ is positive over the entire
range of integration (that is $(0,\pi)$ ), so the integral over $\theta$
does not vanish as it does for the $\cos(\theta)$ in the integral
for the $x$ coordinate.

RonL

**JaysFan31** · August 1st 2006, 07:43 PM

I can't figure out the integral for the y-coordinate. Quick assistance would be greatly appreciated.

Is it the integration of (r^3)sin(theta)sin(theta) drdtheta with the same bounds?

**CaptainBlack** · August 1st 2006, 08:13 PM

Originally Posted by JaysFan31

I can't figure out the integral for the y-coordinate. Quick assistance would be greatly appreciated.

Is it the integration of (r^3)sin(theta)sin(theta) drdtheta with the same bounds?

You want:

$ \bold{R_y}=\frac{1}{M}\int_{\theta=0}^{\pi} \int_{r=6}^7 \kappa\ r^3\ \sin(\theta)\ dr\ d\theta $

RonL

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